You’re given strings Jrepresenting the types of stones that are jewels, and Srepresenting the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.
这题的思路就是同时遍历字符串S和J,如果遍历S的时候发现该字符也出现在J中，则将计数器加1，最后返回计数器的值。
Code:
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
res = 0
for s in S:
if s in J:
res += 1
return res优化点的解决方案是把J转化成set对象，这样lookup的复杂度就减少为O(1)了，参考代码：
class Solution:
def numJewelsInStones(self, J: str, S: str) -> int:
res = 0
for s in S:
if s in set(J):
res += 1
return res